'Series Sums' printed from http://nrich.maths.org/

As each sum develops it should become clear that the last number in each sum is triangular. So for $S_n$, the last number in the sum is the $n^{th}$ triangular number $= n(n + 1)/2$. Bearing this in mind and the fact that the first number in the sum is the $(n - 1)^{th}$ triangular number plus $1$, then,